Two Sum (leetcode 1)

Two Sum Total Accepted: 104422 Total Submissions: 590920 My Submissions Question Solution
Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

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思路：快速向中间推进，

#include <algorithm>#include <functional>vector<int> num;bool cmp(int a, int b) {    return num[a] < num[b];}class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        num = nums;        vector<int> index;        for (int i = 0; i < nums.size(); ++i) {            index.push_back(i);        }        sort(index.begin(), index.end(), cmp);        for(int i = 0; i < index.size(); ++i) {            //cout << nums[index[i]] << " "<<endl;        }        int i = 0;        int j = nums.size() - 1;        vector<int> v;        while(i >= 0 && i < j) {            if (nums[index[i]] + nums[index[j]] == target) {                if (index[i] < index[j]) {                    v.push_back(index[i] + 1);                    v.push_back(index[j] + 1);                } else {                    v.push_back(index[j] + 1);                    v.push_back(index[i] + 1);                }                return v;            }            while (i < j && nums[index[i]] + nums[index[j]] < target) {                ++i;            }            while (i < j && nums[index[i]] + nums[index[j]] > target) {                --j;            }            if (nums[index[i]] + nums[index[j]] < target && i + 1 < j && nums[index[i + 1]] + nums[index[j]] > target) {                j -= 1;            }            if (nums[index[i]] + nums[index[j]] > target && i < j - 1 && nums[index[i]] + nums[index[j - 1]] < target) {                i += 1;            }            // cout << "i:" <<i <<"num:" << nums[index[i]] <<"j:" <<j<<"nums:"<<nums[index[j]]<<endl;        }        return v;    }};

后来从网上看到另一个思路：hash 纪录已经出现的值，而后再直接查找。（用空间换时间，和利用位图法记录出现数据类似。不过在稀疏串中hash更节省内存。